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明治大学 先端数理科学研究科 現象数理学専攻 2019年8月実施 微积分

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Author​

思齐塾, 祭音Myyura

Description​

C2C^2 級の関数 f:(0,∞)→Rf: (0, \infty) \to \mathbb{R} に対して、関数 uu を

u(x,y,z)=f(x2+y2+z2)((x,y,z)≠(0,0,0))u(x, y, z) = f(\sqrt{x^2 + y^2 + z^2}) \quad ((x, y, z) \neq (0, 0, 0))

で定めるとき、以下の問いに答えよ。

(1) ∂u∂x\frac{\partial u}{\partial x} を ff を用いて表せ。

(2) ∂2u∂x2+∂2u∂y2+∂2u∂z2\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2} を ff を用いて表せ。

(3) ∂2u∂x2+∂2u∂y2+∂2u∂z2=0\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2} = 0 が成り立つならば、ある定数 C1,C2C_1, C_2 を用いて

u(x,y,z)=C1x2+y2+z2+C2u(x, y, z) = \frac{C_1}{\sqrt{x^2 + y^2 + z^2}} + C_2

と表されることを示せ。

(4) 原点中心の半径1の球を Ω\Omega とするとき、次の積分の値を計算せよ。

I=∭Ωdxdydzx2+y2+z2I = \iiint_{\Omega} \frac{dx dy dz}{\sqrt{x^2 + y^2 + z^2}}

题目描述​

设 f:(0,∞)→Rf:(0,\infty)\to\mathbb{R} 为 C2C^2 类函数，并对 (x,y,z)≠(0,0,0)(x,y,z)\neq(0,0,0) 定义

u(x,y,z)=f ⁣(x2+y2+z2).u(x,y,z)=f\!\left(\sqrt{x^2+y^2+z^2}\right).

回答下列问题。

(1) 用 ff 表示 ∂u∂x\dfrac{\partial u}{\partial x}。

(2) 用 ff 表示

∂2u∂x2+∂2u∂y2+∂2u∂z2.\frac{\partial^2u}{\partial x^2} +\frac{\partial^2u}{\partial y^2} +\frac{\partial^2u}{\partial z^2}.

(3) 若

∂2u∂x2+∂2u∂y2+∂2u∂z2=0,\frac{\partial^2u}{\partial x^2} +\frac{\partial^2u}{\partial y^2} +\frac{\partial^2u}{\partial z^2}=0,

证明存在常数 C1,C2C_1,C_2，使

u(x,y,z)=C1x2+y2+z2+C2.u(x,y,z)=\frac{C_1}{\sqrt{x^2+y^2+z^2}}+C_2.

(4) 设 Ω\Omega 为以原点为球心、半径为 11 的球体，计算

I=∭Ωdx dy dzx2+y2+z2.I=\iiint_{\Omega}\frac{dx\,dy\,dz}{\sqrt{x^2+y^2+z^2}}.

Kai​

(1) Let r=x2+y2+z2r = \sqrt{x^2 + y^2 + z^2} . Then u(x,y,z)=f(r)u(x, y, z) = f(r) .

∂u∂x=∂f∂r∂r∂x=f′(r)⋅xx2+y2+z2=xrf′(r)\frac{\partial u}{\partial x} = \frac{\partial f}{\partial r} \frac{\partial r}{\partial x} = f'(r) \cdot \frac{x}{\sqrt{x^2 + y^2 + z^2}} = \frac{x}{r}f'(r)

(2)

∂2u∂x2=∂∂x(xrf′(r))=1rf′(r)+x∂∂x(f′(r)r)=1rf′(r)+x(f′′(r)xrr−f′(r)xrr2)=1rf′(r)+x2r2f′′(r)−x2r3f′(r)\frac{\partial^2 u}{\partial x^2} = \frac{\partial}{\partial x} \left( \frac{x}{r} f'(r) \right) = \frac{1}{r} f'(r) + x \frac{\partial}{\partial x} \left( \frac{f'(r)}{r} \right) = \frac{1}{r} f'(r) + x \left( \frac{f''(r) \frac{x}{r} r - f'(r) \frac{x}{r}}{r^2} \right) = \frac{1}{r} f'(r) + \frac{x^2}{r^2} f''(r) - \frac{x^2}{r^3} f'(r)

Similarly,

∂2u∂y2=1rf′(r)+y2r2f′′(r)−y2r3f′(r)\frac{\partial^2 u}{\partial y^2} = \frac{1}{r} f'(r) + \frac{y^2}{r^2} f''(r) - \frac{y^2}{r^3} f'(r)
∂2u∂z2=1rf′(r)+z2r2f′′(r)−z2r3f′(r)\frac{\partial^2 u}{\partial z^2} = \frac{1}{r} f'(r) + \frac{z^2}{r^2} f''(r) - \frac{z^2}{r^3} f'(r)

Therefore,

∂2u∂x2+∂2u∂y2+∂2u∂z2=3rf′(r)+x2+y2+z2r2f′′(r)−x2+y2+z2r3f′(r)=3rf′(r)+f′′(r)−1rf′(r)=f′′(r)+2rf′(r)\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2} = \frac{3}{r} f'(r) + \frac{x^2 + y^2 + z^2}{r^2} f''(r) - \frac{x^2 + y^2 + z^2}{r^3} f'(r) = \frac{3}{r} f'(r) + f''(r) - \frac{1}{r} f'(r) = f''(r) + \frac{2}{r} f'(r)

(3) If ∂2u∂x2+∂2u∂y2+∂2u∂z2=0\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2} = 0 , then f′′(r)+2rf′(r)=0f''(r) + \frac{2}{r} f'(r) = 0 . Let g(r)=f′(r)g(r) = f'(r) . Then g′(r)+2rg(r)=0g'(r) + \frac{2}{r} g(r) = 0 .

(r2g(r))′=r2(g′(r)+2rg(r))=0⇒g(r)=C1r2(r^2g(r))'=r^2\left(g'(r)+\frac{2}{r}g(r)\right)=0\quad\Rightarrow\quad g(r)=\frac{C_1}{r^2}

Thus, f(r)=∫C1r2dr=−C1r+C2f(r) = \int \frac{C_1}{r^2} dr = -\frac{C_1}{r} + C_2 . Therefore, u(x,y,z)=−C1x2+y2+z2+C2=C1′x2+y2+z2+C2u(x, y, z) = -\frac{C_1}{\sqrt{x^2 + y^2 + z^2}} + C_2 = \frac{C_1'}{\sqrt{x^2 + y^2 + z^2}} + C_2 for some constant C1′=−C1C_1' = -C_1 .

(4) In spherical coordinates, x=ρsin⁡ϕcos⁡θx = \rho \sin \phi \cos \theta , y=ρsin⁡ϕsin⁡θy = \rho \sin \phi \sin \theta , z=ρcos⁡ϕz = \rho \cos \phi , where 0≤ρ≤10 \leq \rho \leq 1 , 0≤ϕ≤π0 \leq \phi \leq \pi , 0≤θ≤2π0 \leq \theta \leq 2\pi . Then x2+y2+z2=ρ2x^2 + y^2 + z^2 = \rho^2 , and dxdydz=ρ2sin⁡ϕdρdϕdθdx dy dz = \rho^2 \sin \phi d\rho d\phi d\theta .

I=∭Ωdxdydzx2+y2+z2=∫01∫0π∫02πρ2sin⁡ϕρ2dθdϕdρ=∫01∫0π∫02πρsin⁡ϕdθdϕdρI = \iiint_{\Omega} \frac{dx dy dz}{\sqrt{x^2 + y^2 + z^2}} = \int_0^1 \int_0^{\pi} \int_0^{2\pi} \frac{\rho^2 \sin \phi}{\sqrt{\rho^2}} d\theta d\phi d\rho = \int_0^1 \int_0^{\pi} \int_0^{2\pi} \rho \sin \phi d\theta d\phi d\rho
=∫01ρdρ∫0πsin⁡ϕdϕ∫02πdθ=[ρ22]01⋅[−cos⁡ϕ]0π⋅[θ]02π=12⋅(1+1)⋅2π=2π= \int_0^1 \rho d\rho \int_0^{\pi} \sin \phi d\phi \int_0^{2\pi} d\theta = \left[ \frac{\rho^2}{2} \right]_0^1 \cdot [-\cos \phi]_0^{\pi} \cdot [\theta]_0^{2\pi} = \frac{1}{2} \cdot (1 + 1) \cdot 2\pi = 2\pi
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